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Plotting Kinetic Energy of Mass Against Time and Distance

  1. Nov 1, 2011 #1
    1. The problem statement, all variables and given/known data
    A horizontal force of 80N acts on a mass of 6kg resting on a horizontal face. The mass is initially at rest and covers a distance of 5m in 0.92s under the action of the force. Assuming there are no energy losses due to air resistance and therefore that the acceleration is constant.
    A) Calculate the energy expended in the acceleration.
    B) Plot a graph of the kinetic energy of the mass against time.
    C) Plot a graph of the kinetic energy of the mass against distance.

    2. Relevant equations
    K.E.= (mv^2)/2,
    V= ∆s/∆t
    S= d/t

    3. The attempt at a solution

    A) Work=force ×distance
    W=80 ×5

    B) Speed= distance/time
    S= d/t
    S= 5/0.92

    Velocity= (change in speed)/(change in time)
    V= ∆s/∆t
    V= 5.4348/0.92

    K.E.= (mv^2)/2
    K.E.= (6 × 5.907^2)/2
    K.E.= 104.678J

    If correct, on the x axis i'd have time in s, on the y axis K.E. in Joules. I'd draw a straight line from 0,0 to 0.92,104.678?

    C) Similarly to answer B, except with distance on the x axis and K.E. on the y axis. I'd draw a straight line from 0,0 to 5,104.678?

    If any body is able to check my answers and provide some guidance i'd be most grateful.
    Last edited: Nov 1, 2011
  2. jcsd
  3. Nov 1, 2011 #2

    Simon Bridge

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    looks good to me.
    Not when you are accelerating - your speed is changes with time.
    You need the equation of how speed depends on time.

    [itex]v(t)=at+v_0[/itex] (use the v-t graph for the motion).

    If you put [itex]F=ma[/itex] into [itex]W=Fd[/itex] you get [itex]W=mad[/itex] (you always knew work was mad right?) You are given the mass and the distance, you just calculated the work - so [itex]a=W/md[/itex]. (or just a = F/m but the previous was more entertaining!)

    You should be able to finish from there.

    Note: you won't get very far by just applying equations - you need to understand the physics they describe.
    Last edited: Nov 1, 2011
  4. Nov 1, 2011 #3
    Thanks for your help. So i've got:

    Kinetic Energy= (mass × velocity^2)/2
    K.E.= (mv^2)/2

    Velocity= initial velocity+(acceleration ×time)

    Work=mass ×acceleration ×distance

    a= w/md
    a= 400/(6 × 5)
    a=13.33 ̇m/s

    V=0+(13.33 ×0.92)
    V=12.266 ̇ms^(-1)

    K.E.= (mv^2)/2
    K.E.= (6 ×12.266^2)/2

    With my graphs looking like the attachments. Does this now seem correct?

    Thanks for your help.

    Attached Files:

  5. Nov 1, 2011 #4

    Simon Bridge

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    Should KE change at a constant rate?

    Lets see, v is proportional to t since v(t)=at
    KE is proportional to v-squared

    so how does KE change with time?

    If you still don't see it, computer the kinetic energy at times: t= 0.1s, 0.2s, 0.3s etc and plot them.
    (I could just tell you but if you do this you will learn.)
  6. Nov 1, 2011 #5
    Time v K.E.
    0 0
    0.1 5.07
    0.2 21.32
    0.3 47.98
    0.4 85.29
    0.5 133.27
    0.6 191.9
    0.7 261.2
    0.8 341.16
    0.9 431.78
    0.92 451.36

    So Kinetic Energy will increase exponentially?

    Thanks for you help so far it's been awesome of you.

    Attached Files:

  7. Nov 1, 2011 #6

    Simon Bridge

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    Confirm your conclusion! If it is exponential, then a plot of ln(KE) against time will be a line. Is it?

    What if you plot KE against t2: is that a line?

    You were kind-of expected to reason like this:

    [itex]K=\frac{1}{2}mv^2[/itex] but [itex]v=at[/itex]

    so: [itex]K=\frac{1}{2}ma^2t^2[/itex]

    so: kinetic energy depends on the square of the time.
  8. Jan 8, 2012 #7
    Should the kinetic energy exceed the work done?

    The question states ignore all forces and assume the acceleration is constant. So 451 J seems excessive

    I get the acceleration to equal 12.551 m/s (constant) and velocity at 11.547 @ .92 seconds

    Using k=1/2*m*v^2

    .5*6*11.547^2=399.99 J

    Then use V=at to get the velocity to use at different times I.e 0.1, 0.2 to plot the graph

    This KE
  9. Jan 8, 2012 #8

    Simon Bridge

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    EK gained should not exceed the work done, no.
    But your re-calc'ed EK is 399.99J and the work done is Fd=400J so you are fine.
    The difference is lost to friction with the surface - but that's a very very slippery surface!

    Check your working:
    ... a distance of 5m in 0.92 seconds - constant acceleration.

    a=2d/t2 = 10/(0.92)2 = 11.815ms-2

    ... that looks better.
    I think you used work to compute acceleration - but you forgot friction (only air resistance was ruled out.)

    v=at will give you EK vs t data - correct.

    v2=2ad will give you EK vs d data
  10. Jan 8, 2012 #9
    That makes sense,

    I now get KE = 354.442 J at 0.92 seconds (v=at)


    KE=177.22 J at 5 meters (v^2=ad)
  11. Jan 8, 2012 #10

    Simon Bridge

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    But you have 5m at 0.92s... so they can't both be right?!
    Which one is correct?
  12. Jan 8, 2012 #11
    Pardon my mistake...

    I calculated v^2=ad when I should have used v^2=2ad

    Giving 354.442 J

    Then use the (d) at 1, 2, 3 m etc and implement them into the equation to get my graph points..

    Does 354.442 J seem like it will be the correct answer?
  13. Jan 8, 2012 #12

    Simon Bridge

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    Well if you look at it, it is less than the work, which is good.
    The difference is the energy lost to friction.

    Also you did it two different ways and got the same answer.
    Which should give you confidence.

    Use points closer together. I use gnu octave:

    Code (Text):

    t=0:0.02:0.92; %prepares 46 points
    d=0:0.125:5; %prepares 40 points


    title{"Kinetic Energy vs time"}

    title("Kinetic energy vs distance")
  14. Jan 8, 2012 #13
    Thanks for the great help Simon,I'll get on with them graphs
  15. Jan 8, 2012 #14
    Calculate the coefficient of friction between the mass and the surface

    a = 11.815 m/s2

    Fn = m*a Difference between 80N and the friction Ff

    m*a = 6*11.815 = 70.89 N

    Ff = 80 – 70.89 = 9.11 N

    Ff = mu*m*g

    mu = Ff / (m*g)

    9.11/(6*9.81)= 0.154
  16. Jan 8, 2012 #15

    Simon Bridge

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    Using conservation of energy is more elegant and gets you there faster:
    Total work is equal to the work done to accelerate the mass plus that done against friction. Thus, work against friction = 400-354.44=45.56J = Ff.d

    Thus Ff = 45.56/5 = 9.11N ... just the same.

    You're getting the hang of this :)
  17. Jan 12, 2012 #16
    so what do the graphs look like?
  18. Jan 21, 2013 #17
    Did anyone publish the plotted graphs for this..?
  19. Jan 31, 2013 #18
    How did you get acceleration to equal 12.551 m/s?

    Using w/md=a I got 13.33 m/s which ended up giving me 451.19J @ 0.92s which has already been pointed out to be too high a value for the KE. I can't see where I'm going wrong??
  20. Mar 20, 2013 #19
    Anyone got the solution to this question for calculating the acceleration as 12.551?
    Which formula was used.
  21. Mar 21, 2013 #20
    Anyone able to help with the correct acceleration formula for this problem, like Carlo, i used newtons second law which ended up giving me 451.19J, more than the work done value so incorrect.
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